Printable · GCSE Higher · ages 14-16
Number worksheet — GCSE Higher
Fifteen questions across the number statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Number worksheet — GCSE Higher
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- 1.Write these three numbers in order of size, starting with the smallest: 0.7, 3/4, 0.72
- 2.Round 0.006482 to 2 significant figures.
- 3.Write down the reciprocal of 0.2
- 4.Write 200 as a product of its prime factors, using index notation.
- 5.Simplify √45.
- 6.p = 24 and q = 6, each correct to the nearest integer. Work out the greatest degree of accuracy to which p ÷ q can be guaranteed correct.
- 7.A recipe uses 160 g of flour. Sam wants to increase the amount by 1/4. Work out the new amount of flour.
- 8.Work out 6² − 4².
- 9.A spreadsheet shows that 812 − 397 = 315. Work out an estimate for 812 − 397, by rounding each number to the nearest 100, to check whether the spreadsheet's answer is reasonable.
- 10.Write 7/12 as a decimal, showing clearly which digit is recurring.
- 11.Work out 100 − 4 × 5²
- 12.A student works out the exact area of a circle with radius 4 cm by squaring the radius but forgetting to multiply by π. Work out the correct exact area of the circle, in terms of π.
- 13.Amelia estimates 48 × 21 by working out 50 × 20 = 1,000. Work out whether her estimate is an under-estimate or an over-estimate, and by how much.
- 14.Work out √25 + 4² − 12 ÷ 3
- 15.Expand and simplify (2 + √3)², giving your answer in the form a + b√3.
Answer key
- (c) 0.7, 0.72, 3/4 — Method: numbers written in different forms cannot be compared as they stand, so every fraction is turned into a decimal by dividing the numerator by the denominator, and the decimals are then compared place by place from the left. Working: 3/4 means 3 ÷ 4 = 0.75, so the three values to compare are 0.7, 0.75 and 0.72; written to two decimal places they are 0.70, 0.75 and 0.72, and the hundredths digits 0, 5 and 2 put 0.70 first, 0.72 next and 0.75 last; written again in the forms the question used, the order from smallest is 0.7, then 0.72, then 3/4. Answer: 0.7, 0.72, 3/4. The distractors: 3/4, 0.7, 0.72 comes from turning 3/4 into 0.34 by writing the numerator and the denominator as the two digits after the point, which makes the fraction the smallest of the three; 0.72, 3/4, 0.7 comes from the belief that the more digits a decimal has the smaller it must be, which puts both 0.72 and 0.75 below 0.7 and 0.72 below 0.75; 3/4, 0.72, 0.7 comes from comparing the three values correctly but listing them largest first, against an instruction to start with the smallest.
- (c) 0.0065 — Leading zeros are never significant, so counting from the first non-zero digit, the first two significant figures of 0.006482 are 6 and 4. Look at the next digit along, 8, to decide whether the second figure rounds up: since 8 is 5 or more, the 4 rounds up to 5, giving 0.0065. Rounding to 2 decimal places instead of 2 significant figures gives 0.01, which answers a different question. Wrongly counting one of the leading zeros as a significant figure and stopping one figure short gives 0.006. Keeping an extra digit, as in 0.00648, gives 3 significant figures rather than 2.
- (c) 5 — 0.2 = 1/5, and turning the fraction upside down gives the reciprocal 5/1 = 5. Writing −0.2 mistakes the reciprocal for the negative of the number. Working out 1 − 0.2 = 0.8 mistakes the reciprocal for subtracting the number from 1. Writing 0.5 is the reciprocal of 2, not of 0.2 — a place-value slip that drops the decimal, since 1 ÷ 0.2 = 5 while 1 ÷ 2 = 0.5.
- (d) 2³ × 5² — Method: divide repeatedly by the smallest prime number, then write any repeated prime using a power. Working: 200 ÷ 2 = 100, 100 ÷ 2 = 50, 50 ÷ 2 = 25, 25 ÷ 5 = 5, and 5 is prime, so 200 = 2 × 2 × 2 × 5 × 5, written as 2³ × 5². 2² × 5³ swaps the two powers, giving 4 × 125 = 500, not 200. 2³ × 5 leaves out one of the two 5s, giving 8 × 5 = 40, not 200. 2 × 5³ leaves out two of the three 2s, giving 2 × 125 = 250, not 200. Answer: 2³ × 5².
- (d) 3√5 — Split 45 into a perfect square times a factor: 45 = 9 × 5. Take the square root of each part separately: √45 = √9 × √5 = 3√5, since √9 = 3. Writing the perfect-square factor itself (9) as the coefficient instead of its root would give 9√5 — that trap comes from forgetting the last step, rooting 9. Multiplying 3 and 5 together instead of keeping them as coefficient and radicand gives 15, which throws away the surd entirely. Doubling the correct coefficient by mistake gives 6√5.
- (a) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
- (a) 200 g — One quarter of 160 g is 40 g. Increasing the amount means adding this on: 160 + 40 = 200 g. Finding the increase, 1/4 of 160 = 40 g, but stopping there without adding it to the original amount leaves just 40 g. Using 4/5 instead of 5/4 as the scaling fraction, 160 × 4/5 = 128 g, actually decreases the amount rather than increasing it. Increasing by a half instead of a quarter, 160 + 80 = 240 g, uses the wrong fraction of 160.
- (c) 20 — Method: work out each power separately before subtracting. Working: 6² = 36 and 4² = 16, so 6² − 4² = 36 − 16 = 20. Answer: 20. (4 comes from subtracting first, 6 − 4 = 2, and then squaring that result, instead of squaring each number first. 52 comes from adding the two squares, 36 + 16, instead of subtracting them. 2 comes from subtracting the two numbers, 6 − 4, and forgetting to square at all.)
- (b) 400 — Method: round each number to the nearest 100, then subtract the rounded values. Working: 812 rounds to 800 (nearest 100) and 397 rounds to 400 (nearest 100). 800 − 400 = 400. Answer: 400. 500 comes from rounding 397 down to 300 instead of up to the nearest 100, 400. 300 comes from rounding 812 down to 700 instead of up to the nearest 100, 800. 415 is the exact value of 812 − 397, found without rounding first, so it is not an estimate — the spreadsheet's answer of 315 is too far from the estimate of 400 to be correct.
- (b) 0.58333... — Divide 7 by 12 using long division. 70 ÷ 12 = 5 remainder 10, so the first decimal digit is 5. Bring down a 0 to make 100: 100 ÷ 12 = 8 remainder 4, so the second digit is 8. Bring down a 0 to make 40: 40 ÷ 12 = 3 remainder 4, so the third digit is 3. Bring down a 0 to make 40 again — the remainder 4 has reappeared, so the digit 3 repeats forever from here. This gives 7/12 = 0.58333... . Stopping the division after two digits and writing 0.58 treats it as if it terminated, when the remainder is not yet zero. Misreading the pattern as a two-digit repeating block, '58', gives 0.585858..., which wrongly makes the 5 recur as well as the 3. A slip in the long division that carries the wrong remainder forward can make the second digit itself appear to repeat instead of the third, giving 0.588888... .
- (a) 0 — Method: BIDMAS works through the index first, then the multiplication, then the subtraction. Working: 5² = 25, then 4 × 25 = 100, and finally 100 − 100 = 0. Answer: 0. The distractors: 2400 comes from working from left to right and subtracting first, giving (100 − 4) × 25 = 96 × 25 = 2400; −300 comes from multiplying before applying the index, giving (4 × 5)² = 20² = 400 and then 100 − 400 = −300; 60 comes from reading 5² as 5 × 2 = 10, so that 4 × 10 = 40 and 100 − 40 = 60.
- (d) 16π cm² — Method: for a circle, area = π × radius². Working: area = π × 4² = π × 16 = 16π cm². Answer: 16π cm². The student squared the radius but left out the π, which is why 16 cm² is not the exact area. 4π cm² comes from multiplying by the radius once instead of squaring it: π × 4 = 4π. 8π cm² comes from using the circumference formula 2 × π × radius instead of the area formula: 2 × π × 4 = 8π. 64π cm² comes from using the diameter (8 cm) as the radius in the area formula: π × 8² = 64π.
- (c) An under-estimate, by 8 — Method: work out the exact product, then compare it with the estimate; an estimate that is smaller than the exact value is an under-estimate, and the difference between them is the size of the error. Working: 48 × 21 = 48 × 20 + 48 = 960 + 48 = 1,008, and 1,008 − 1,000 = 8, so the estimate falls short. Answer: an under-estimate, by 8. The distractors: an over-estimate by 8 has the size of the error right but the direction wrong, and comes from assuming that rounding 48 up to 50 must push the estimate above the exact value, without allowing for 21 being rounded down; an over-estimate by 19 comes from working out 48 × 21 as 48 × 20 + 21 = 981, adding a 21 where another 48 belongs; the claim that the estimate is exactly right comes from arguing that one number was rounded up and the other down, so the two changes must cancel.
- (b) 17 — Roots and powers are worked out first: √25 = 5 and 4² = 16. Division comes next: 12 ÷ 3 = 4. Then addition and subtraction, left to right: 5 + 16 − 4 = 17. A candidate who treated 4² as 4 × 2 = 8, multiplying the base by the exponent instead of squaring it, worked out 5 + 8 − 4 = 9. A candidate who did not evaluate the root and used 25 itself worked out 25 + 16 − 4 = 37. A candidate who ignored the priority of division and worked through 5 + 16 − 12 ÷ 3 strictly left to right got 5 + 16 = 21, then 21 − 12 = 9, then 9 ÷ 3 = 3.
- (c) 7 + 4√3 — Expand the brackets fully: (2 + √3)² = 2² + 2 × 2 × √3 + (√3)² = 4 + 4√3 + 3. Adding the two whole-number terms, 4 + 3 = 7, gives 7 + 4√3. Using (a + b)² = a² + b² and skipping the middle cross term entirely gives just 4 + 3 = 7, with no surd term at all. Treating (√3)² as if it stayed √3 rather than becoming 3, then merging it with the existing surd term, gives 4 + 5√3. Squaring only the surd term correctly but carrying the whole-number term as 2 instead of squaring it to 4 gives 2 + 3 + 4√3 = 5 + 4√3.
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